For a transparent prism, if the angle of minimum deviation is equal to its refracting angle, the refractive index n of the prism satisfies

Q. For a transparent prism, if the angle of minimum deviation is equal to its refracting angle, the refractive index n of the prism satisfies.

A. n ≥ 2

B. √2 < n < 2

C. 1 < n < 2

D. √2 < n < 2√2

Correct Answer: √2 < n < 2

Solution

For a prism, the relation between refractive index n, angle of prism A, and angle of minimum deviation δ_m is:

n = sin[(A + δ_m)/2] / sin(A/2)

Given that the angle of minimum deviation is equal to the refracting angle:

δ_m = A

Substituting in the formula:

n = sin(A) / sin(A/2)

Using the identity:

sin A = 2 sin(A/2) cos(A/2)

We get:

n = 2 cos(A/2)

For a prism, the refracting angle satisfies:

0 < A < π/2

Hence:

π/4 < A/2 < π/2

This gives the range:

√2 < n < 2

Therefore, the correct option is √2 < n < 2.